For the second one it's quite easy actually... it's 0.

You have thousands, hundreds, tens and no digit for unit's so it's 0.

fox_a_lisa wrote:Hi all,

Can you please give me the explanation of this two questions:

1. How many integers n greater then 10 and less then 100 are these such that, if the digits of n are reversed, the resulting integer is n-9?
A) 5
B) 6
C) 7
D) 8
E) 9

Let n integer is 10x+y
then from given condition

10x+y - 9 = 10y+x

or 9x-9y = 9
x-y = 1

so all the 2 digit numbers wherein difference between the digit is 1 satisfy the condition

therefore
21
32
43
54
65
76
87
98

IMO D

Great! Thank you!!

I did #1 a little more brute force, but still solved it pretty quickly.

the first number where the digits can be reversed is 12, whcich reverses to 21. Those two are 9 apart.

Then 13 -> 31, the spread is too great and the spread will continue to get larger for the rest of the teens.

Any number ending in 0 will have too great a spread, so 20,30,40,50, 60,70, 80 and 90 you don't need to worry about.

We have alread looked at 21 (with 12), 22 doesn't reverse to a new number, so we go to 23->32, which are 9 apart.

The pattern emerged quickly that there would be one pair for each 10's There are 9 10s, so 9 pairs that satisfy the equation.

Answer is E.

-Carrie

The Answer is D.

The numbers are in the form of x(x+1) ie 12, 23

The possible of x are x = 1 to x+1 = 9 ie x :{1,2....8}

So there are 8 numbers between 10 and 100.

Carrie Note that 89 is the last number below 100. For each 10 numbers you have one such number but between 90 and 100 you don't have a number whose reverse is below 100.

it should be 90-10/10 = 8